Solving Polar Co-ordinate Paradox: Finding \frac {\partial r} {\partial x}

[dirtydog]

Hi I am having a bit of difficulty working with plane polar co-ordinates.
We have:
[tex] r^2 = x^2 + z^2 [/tex]
[tex] x = r cos \theta [/tex]
[tex] z=r sin \theta [/tex]
I wish to find [tex] \frac {\partial r} {\partial x} [/tex]

Using [tex] r^2 = x^2 + z^2 [/tex]
We have:

[tex] \frac {\partial (r^2)} {\partial x} = \frac {\partial (x^2)} {\partial x} + \frac {\partial (z^2)} {\partial x} [/tex]
Thus [tex] 2r\frac {\partial r} {\partial x} = 2x [/tex]
[tex] \frac {\partial r} {\partial x} = \frac {x} {r} = \frac {r cos \theta} {r} [/tex]
Therefore [tex]\frac {\partial r} {\partial x} = cos \theta [/tex]

But if we find [tex] \frac {\partial r} {\partial x} [/tex] using [tex] x = r cos \theta [/tex]
We have:
[tex] r = \frac {x} {cos \theta} [/tex]
Therefore [tex]\frac {\partial r} {\partial x} = \frac {1} {cos \theta} [/tex]

What is going on here? Which answer is wrong and why?

 

[Tide]

Ooops! In your last step you [itex]\theta[/itex] is NOT a constant! You need to differentiate it wrt x as well.

 

[shmoe]

But if we find [tex] \frac {\partial r} {\partial x} [/tex] using [tex] x = r cos \theta [/tex]
We have:
[tex] r = \frac {x} {cos \theta} [/tex]
Therefore [tex]\frac {\partial r} {\partial x} = \frac {1} {cos \theta} [/tex]

Hi, when you differentiated with respect to x here, you treated [tex]\theta[/tex] as a constant. It's not, it's dependant on x, so you have to use the quotient rule and you'll have a [tex]\frac {\partial \theta} {\partial x}[/tex] appear via the chain rule. Use [tex]{\cos \theta}=\frac{x}{\sqrt{x^2+z^2}}[/tex] to work out [tex]\frac {\partial \theta} {\partial x}[/tex].

eidt-Tide was quicker on the draw!

 

[NeutronStar]

I'm not sure, but I think your mistake is in the following procedure:

[tex]r = \frac {x} {cos \theta} \rightarrow \frac {\partial r} {\partial x} = \frac {1} {cos \theta} [/tex]

I believe that [tex] \theta [/tex] is dependent on [tex] x [/tex]. In other words [tex] \theta = \theta (x)[/tex] so you can't just treat [tex] cos \theta[/tex] as a constant when taking the derivative with respect to [tex] x [/tex].

 

[NeutronStar]

Whoops! I guess I'm pretty late on that one!

This is the first time I used the equation formatting and it took me a while to format the post. (ha ha)

At least, I know now that I was on the right track.

 

[Tide]

Way to go, NS!

 

[dirtydog]

Thanks for your help guys!